Problem 5

Line

\ell

intersects sides

BC

and

AD

of cyclic quadrilateral

ABCD

at its interior points

R

and

S

, respectively, and intersects ray

DC

beyond point

C

Q

, and ray

BA

beyond point

A

P

. The circumcircles of triangles

QCR

and

QDS

intersect at

N\neq Q

, while the circumcircles of triangles

PAS

and

PBR

intersect at

M\neq P

. Let lines

MP

and

NQ

meet at point

X

, lines

AB

and

CD

meet at point

K

, and lines

BC

and

AD

meet at point

L

. Prove that point

X

lies on line

KL

Step 6 of 6: Conclude X lies on KL

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X=MP\cap NQ=T=V\in KL

Detailed analysis

Since $T=NQ\cap KL$ and $V=MP\cap KL$ coincide, this common point lies on both line $NQ$ and line $MP$ , hence equals $X=MP\cap NQ$ . Because $T=V$ lies on $KL$ by construction, $X$ lies on line $KL$ , as required.